Welcome to MathOverflow

Question

2

1

I'm not sure how long this iterative questions can go on, but let me try again. Let's say $X$ is a Cohen-Macaulay scheme with an action of $\mathbb{G}_m$ (i.e. if $X$ is affine, a grading on the coordinate ring). Are the schematic fixed points $X^{\mathbb{G}_m}$ of $X$ Cohen-Macaulay?

edited Oct 25 2010 at 23:02

asked Oct 25 2010 at 21:30

Ben Webster♦
20.7k●1●44●114

2

I am not sure what the usual business is, but it is not true that if $\mathbb G_{\rm m}$ acts on a variety $X$ with a fixed point $p$, then this induces an action of $\mathbb G_{\rm m}$ on $\mathop{\rm Spec}\mathcal O_{X,p}$. – Angelo Oct 25 2010 at 21:39

Is this an issue of not being able to find a $\mathbb{G}_m$-invariant affine open containing $p$? – Ben Webster♦ Oct 25 2010 at 22:23

Those invariant affines are not cofinal in all neighborhoods. Algebraically, the localization of C[x] at the ideal (x) does not admit a grading...does it? – David Treumann Oct 25 2010 at 22:34

Ah, right. That was complete nonsense. Removed. I think you probably you can reduce to the graded local case, but let me not worry about that. – Ben Webster♦ Oct 25 2010 at 23:03

Let $T^1$ act on ${\mathbb A}^3$ with weights $0,1,1$. Then on the ${\mathbb P}^2$, the fixed-point set is not equidimensional. So you can only hope to have a local statement. – Allen Knutson Oct 26 2010 at 0:39

show 1 more comment

Angelo · Accepted Answer · 2010-10-27 05:55:40Z UTC

Here is a counterexample. Consider the action of $\mathbb G_{\rm m}$ on $\mathbb A^4$ defined by $t \cdot(x,y,z,w) = (x, y, tz, t^{-1}w)$, and let $X$ be the invariant closed subscheme with ideal $(xy, y^2 + zw)$; this is a complete intersection, hence it is Cohen-Macaulay. The fixed point subscheme is obtained by intersecting with the fixed point subscheme in $\mathbb A^4$, which is given by $z = w = 0$; hence it is the subscheme of $\mathbb A^2$ given by $xy = y^2 = 0$, which is of course the canonical example of a non Cohen-Macaulay scheme.

Developing this idea a little, one can show that any kind of horrible singularity can appear in the fixed point subscheme of a $\mathbb G_{\rm m}$-action on a complete intersection variety.

Hailong Dao · Answer 2 · 2010-10-26 05:48:51Z UTC

3

Edit: the following does not answer Ben's question. It gives an example of the subring fixed by $G_m$ being not CM, while the question asked about the subscheme of fixed points, see the comments for more details.

Let $R$ be the (homogenous) cone of a curve $C$ of genus $g>0$, for example $R=\mathbb C[x,y,z]/(x^3+y^3+z^3)$. Let $S=R[u,v]$, $X=\text{Spec}(S)$ and $G_m$ acts by

$a.(x,y,z,u,v) = (ax,ay,az,a^{-1}u, a^{-1}v)$.

Then $A= S^{G_m}$ would be a homogenous coordinate ring for $Y= C\times \mathbb P^1$, so it is not Cohen-Macaulay (if $A$ is CM, it would mean that $H^1(Y,\mathcal O_Y)=0$, impossible, see here for an explanation).

(I learned this idea from Hochster, let me try to find a reference)

edited Oct 26 2010 at 5:48

answered Oct 25 2010 at 22:32

Hailong Dao
15.4k●29●69

Hailong- this is the categorical quotient, not schematic fixed points. The schematic fixed points here are a single point with reduced structure and thus Cohen-Macaulay. – Ben Webster♦ Oct 25 2010 at 23:12

Hmm, sorry, I always thought of this notation as the invariants. What do you mean by schematic fixed pts? Are they just literally the pts of $X$ fixed by the group action? – Hailong Dao Oct 26 2010 at 0:10

@Hailong: you applied the fixed-point functor to the algebra rather than to the space. That is, the notation itself has a consistent meaning. (I'm not surely how to reasonable define schematic fixed points; in characteristic zero, with a connected group, one could look where the generating vector fields vanish.) – Allen Knutson Oct 26 2010 at 0:43

Dear Hailong: No ad hoc constructions/definitions are required. For any scheme $X$ over a ring $k$ and action on $X$ by a $k$-gp scheme $G$, define functor $X^G$ on $k$-algebras as follows: for $k$-algebra $R$, $X^G(R)$ is set of $x \in X(R)$ fixed by the $G_R$-action on $X_R$ (i.e., for any $R$-algebra $R'$, $x$ viewed in $X(R')$ is $G(R')$-invariant). Is this represented by a closed subscheme of $X$? Yes, provided $X$ is locally of finite type and separated over $k$ and $G$ is affine and fppf over $k$ with connected fibers. (See Prop. A.8.10(1)ff. in "Pseudo-reductive groups" for details.) – BCnrd Oct 26 2010 at 1:28

1

@ Allen and BCnrd: thank you! I will leave my answer, so people might benefit from your explanations and avoid my mistakes! – Hailong Dao Oct 26 2010 at 2:07

show 1 more comment

Are schematic fixed-points of a Cohen-Macaulay scheme Cohen-Macaulay?

Remember to vote up questions/answers you find interesting or helpful (requires 15 reputation points)

2 Answers

You can accept an answer to one of your own questions by clicking the check mark next to it. This awards 15 reputation points to the person who answered and 2 reputation points to you.